Question

'y varies inversely as x² and y = 8 when x =2 then find

(i) The value of y when x = 4
(ii) What is the effect on the value of y if the value of x is doubled?

Answer 1

$$\begin{gathered} y\alpha \frac{1}{x^2} \\ \Rightarrow y{x}^2=k \\ \Rightarrow y=\frac{k}{x^2}...\left(1\right) \\ \Rightarrow 8\times {\left(2\right)}^2=k(y=8\mathrm{when}x=2.) \\ \Rightarrow k=32 \\ \therefore y{x}^2=32(\mathrm{Equation}\mathrm{of}\mathrm{variation}) \\ \left(\mathrm{i}\right)y{\left(4\right)}^2=32(\because x=4) \\ \therefore y=\frac{32}{16}=2 \\ \left(\mathrm{ii}\right)\mathrm{Let}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{be}2x\mathrm{and}\mathrm{the}\mathrm{new}\mathrm{value}\mathrm{of}\mathrm{y}\mathrm{be}{y}_1. \\ \mathrm{Putting}\mathrm{the}\mathrm{final}\mathrm{value}\mathrm{of}x\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}: \\ {y}_1(2x{)}^2=k \\ \Rightarrow 4y_1{x}^2=k \\ \Rightarrow y_1=\frac{k}{4x^2}...(2) \\ \mathrm{From}(1)\&(2),\mathrm{we}\mathrm{get}: \\ {y}_1=\frac{1}{4}y \end{gathered}$$